What is the integral of (sec^6(θ))/(tan^2(θ)) ?

The integral of (sec^6(θ))/(tan^2(θ)) is -cot(θ)+2tan(θ)+1/3 tan^3(θ)+C

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integral of (sec^6(θ))/(tan^2(θ))

\int \frac{sec ^{6} ( θ )}{tan ^{2} ( θ )} d θ

- cot (θ) + 2 tan (θ) + \frac{1}{3} tan^{3} (θ) + C

Solution steps

\int \frac{sec ^{6} ( θ )}{tan ^{2} ( θ )} d θ

Rewrite using trig identities

= \int \frac{1}{cos ^{4} ( θ ) sin ^{2} ( θ )} d θ

Apply Trigonometric Substitution

: \int \frac{( 1 + u ^{2} ) ^{2}}{u ^{2}} d u

= \int \frac{( 1 + u ^{2} ) ^{2}}{u ^{2}} d u

Expand

\frac{( 1 + u ^{2} ) ^{2}}{u ^{2}} : \frac{1}{u ^{2}} + 2 + u^{2}

= \int \frac{1}{u ^{2}} + 2 + u^{2} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{1}{u ^{2}} d u + \int 2 d u + \int u^{2} d u

\int \frac{1}{u ^{2}} d u = - \frac{1}{u}

\int 2 d u = 2 u

\int u^{2} d u = \frac{u ^{3}}{3}

= - \frac{1}{u} + 2 u + \frac{u ^{3}}{3}

Substitute back

u = tan (θ)

= - \frac{1}{tan ( θ )} + 2 tan (θ) + \frac{tan ^{3} ( θ )}{3}

Simplify

- \frac{1}{tan ( θ )} + 2 tan (θ) + \frac{tan ^{3} ( θ )}{3} : - cot (θ) + 2 tan (θ) + \frac{1}{3} tan^{3} (θ)

= - cot (θ) + 2 tan (θ) + \frac{1}{3} tan^{3} (θ)

Add a constant to the solution

= - cot (θ) + 2 tan (θ) + \frac{1}{3} tan^{3} (θ) + C

\int a \cdot e^{- a x} d x

\int \frac{1}{( cos ^{2} ( x ) 1 + tan ( x ) )} d x

\int \frac{arctan ^{3} ( x )}{3 x ^{2} + 3} d x

\int arctan (tan (t)) d t

What is the integral of (sec^6(θ))/(tan^2(θ)) ?
The integral of (sec^6(θ))/(tan^2(θ)) is -cot(θ)+2tan(θ)+1/3 tan^3(θ)+C